A gas undergoes a polytropic process (PV1.2 = constant) from an initial state of 3.4 bar and 0.283 m3 to a final pressure of 8.2 bar. If 0.183 kg of gas is involved and 2.2 kJ of heat is transferred f

Given:

• Initial pressure, $p_1 = 3.4 \text{ bar} = 3.4 \times 100 \text{ kPa} = 340 \text{ kPa}$
• Initial volume, $V_1 = 0.283 \text{ m}^3$
• Final pressure, $p_2 = 8.2 \text{ bar} = 8.2 \times 100 \text{ kPa} = 820 \text{ kPa}$
• Polytropic index, $n = 1.2$ (from $PV^{1.2} = \text{constant}$)
• Mass of gas, $m = 0.183 \text{ kg}$
• Heat transfer from the gas, $Q = -2.2 \text{ kJ}$ (negative sign indicates heat leaving the system)
• Changes in kinetic and potential energy are negligible.

For a polytropic process, $p_1 V_1^n = p_2 V_2^n$.

$$V_2 = V_1 \left(\frac{p_1}{p_2}\right)^{1/n}$$

Substitute the values:

$$V_2 = 0.283 \text{ m}^3 \left(\frac{3.4 \text{ bar}}{8.2 \text{ bar}}\right)^{1/1.2}$$ $$V_2 = 0.283 \text{ m}^3 \left(\frac{3.4}{8.2}\right)^{0.8333}$$ $$V_2 = 0.283 \text{ m}^3 (0.41463)^{0.8333}$$ $$V_2 = 0.283 \text{ m}^3 \times 0.50611$$ $$V_2 = 0.14332 \text{ m}^3$$

For a polytropic process with $n \neq 1$, the work done is given by:

$$W = \frac{p_2 V_2 - p_1 V_1}{1-n}$$

Substitute the values:

$$W = \frac{(820 \text{ kPa})(0.14332 \text{ m}^3) - (340 \text{ kPa})(0.283 \text{ m}^3)}{1 - 1.2}$$ $$W = \frac{117.5224 \text{ kJ} - 96.22 \text{ kJ}}{-0.2}$$ $$W = \frac{21.3024 \text{ kJ}}{-0.2}$$ $$W = -106.512 \text{ kJ}$$

The negative sign indicates that work is done on the gas (compression).

The First Law of Thermodynamics for a closed system with negligible changes in kinetic and potential energy is:

$$\Delta U = Q - W$$

Substitute the values for $Q$ and $W$:

$$\Delta U = (-2.2 \text{ kJ}) - (-106.512 \text{ kJ})$$ $$\Delta U = -2.2 \text{ kJ} + 106.512 \text{ kJ}$$ $$\Delta U = 104.312 \text{ kJ}$$

Specific internal energy is the total internal energy divided by the mass:

$$\Delta u = \frac{\Delta U}{m}$$

Substitute the values:

$$\Delta u = \frac{104.312 \text{ kJ}}{0.183 \text{ kg}}$$ $$\Delta u = 569.99 \text{ kJ/kg}$$

The heat transfer is given in the problem statement.

The change in specific internal energy of the gas is $\boxed{\text{570.00 kJ/kg}}$.
The heat transfer is $\boxed{\text{-2.2 kJ}}$.